Question

对于满足p在[0,4]范围内所有实数p,使不等式x^2+px>4x+p-3都成立的x的取值范围是多少?

Answer 1

把P=0代入
x2-4x+3＞0即（x-1）（x-3）＞0，解得x＞3或x＜1 ①
把P=4代入不等式得x2-1＞0即（x+1）（x-1）＞0，解得x＞1或x＜-1 ②
解得x＞3或x＜-1，
x（-∞，-1）∪（3，+∞）

Answer 2

原式化为x方+(p-4)x-(p-3)>0 (x+p-3)(x-1)>0 x1=3-p x2=1.又0<=p<=4, -1<=x1<=3 当-1<=x1<1 解为x<-1或x>1 当1<x1<=3,解为x<1或x>3 综合得x<-1或x>3